Theorem 14 (Gnedenko (1943), pp. 442-445)

n[1-F(x)] converges to a W-function of the type W₂(a,x) as n approaches infinity (or, what is the same, Fⁿ(x) converges to a frequency distribution of the type $\psi_a(x)$) if and only if

(a)

there exists an x₀ such that for all $\epsilon \gt 0$,

$$F(x_0) = 1 \quad \text{and} \quad F (x_0-\epsilon) < 1.$$

(b)

for all k > 0 and some a > 0

$$\lim_{x\to-0} \left( \frac{1 - F (kx+x_0)}{1-F(x+x_0)} \right) = k^a.$$

Proof: Assume first that conditions (a) and (b) hold. It will be demonstrated that n[1-F(x)] converges to a W-function of the type W₂ (a,x). Define a_n to be the least upper bound of the x > 0, which satisfy

 

$$1 - F (-x(1-0) + x_0) \leq \frac1n \leq 1 - F (x (1+0)+ x_0),$$ (65)

where n is a positive integer, n > 1. Then, a_n is the greatest lower bound of the values of -x + x₀ which satisfy the inequality, and by Theorem 12, a_n also satisfies the inequality. As a consequence of condition (a) and the properties of frequency distributions, only values of x > 0 can satisfy Eq. (65). Hence, a_n > 0.

By condition (a), as $n \to \infty$, $a_n \to 0$. By condition (b), if x is replaced by $-a_n (1+\epsilon)$ and $k = -x/(1+\epsilon)$, then

 

$$\lim_{-a_n(1+\epsilon)\to-0} \left[ \frac{1 - F (a_n x + x_0... ...n (a+\epsilon) + x_0)} = \left( \frac{x}{1+\epsilon} \right)^a.$$ (66)

Similarly, if x is replaced by $-a_n (a-\epsilon)$ in condition (b) and $k = -x/(1-\epsilon)$, then the condition becomes

 

$$\lim_{-a_n(1-\epsilon)\to-0} \left[ \frac{1-F (a_n x + x_0)}... ...(1-\epsilon) + x_0)} = \left( - \frac{x}{1-\epsilon} \right)^a.$$ (67)

In both Eqs. (66) and (67), it is assumed that $\epsilon \gt 0$ and that $a_n (1-\epsilon) \gt 0$. The left-hand members of Eqs. (66) and (67) are monotonic functions of $\epsilon$ while the right-hand members are continuous functions of $\epsilon$. Hence, by Theorem 10 the convergence in Eqs. (66) and (67) is uniform for $0 \leq \epsilon \leq 1/2$, and

 

$$\begin{array} {r@{~}c@{~}l} \displaystyle \lim_{n\to\infty} ... ... \left( - \frac{x}{1 + \epsilon} \right)^a = (-x)^a,\end{array}$$ (68)

 

$$\begin{array} {r@{~}c@{~}l} \displaystyle \lim_{n\to\infty} ... ...\left( - \frac{x}{1 - \epsilon} \right)^a = (-x)^a. \end{array}$$ (69)

Now, by Eq. (65),

$$\frac{1 - F (a_n x + x_0)}{1 - F (-a_n (1+0) + x_0)} \leq n ... ..._0)] \leq \frac{1 - F (a_n x + x_0)}{1 - F (-a_n (1-0) + x_0)}.$$

Passing to the limit as n,

$$(-x)^a \leq \lim_{n\to\infty} n [ 1 - F (a_n x + x_0)] \leq (-x)^a$$

or

$$\lim_{n\to\infty} n [ 1 - F (a_n x + x_0)] = (-x)^a.$$

Hence, n [1-F (x)] converges to a W-function of the type W₂ (a,x) for x < x₀. By condition (a), F(x) = 1 for $x\geq x_0$,and so for

$$F (a_n x + x_0) = 1, \quad \text{for } \left\{ \begin{aligne... ... x_0, \\ a_n x & \geq 0, \\ x & \geq 0, \end{aligned}\right.$$

or for $x \geq 0$,

$$\lim_{n\to\infty} n [ 1 - F (a_n x + x_0)] = 0.$$

Thus, n [ 1 - F (x)] converges to a W-function of the type W₂ (a,x) for every value of x.

Now, suppose that n [1-F(x)] converges to a W-function of the type W₂ (a,x). Then, for all values of x and for some set of constants a_n > 0 and b_n, $n = 1,2,3,\ldots$,

$$\lim_{n\to\infty} n [1-F (a_n x + b_n)] = W_2 (a,x).$$

Then, by Theorem 7 and Corollary 4 for all x,

 

$$\psi_a (x) = \lim_{n\to\infty} F^n (a_n x + b_n) = \begin{ca... ...{ and some } a \gt 0, \\ 1, & \text{for } x \geq 0. \end{cases}$$ (70)

For the subsequence $\{ 2k\}$, $k = 1,2,3, \ldots$ of $\{ n\}$, the limit of Eq. (70) still exists and

 

$$\lim_{n\to\infty} F^{2n} (a_{2n} x + b_{2n} ) = \left. \begi... ...\\ 1, & \text{for } x \geq 0\end{cases} \right\} = \psi_a (x).$$ (71)

Multiplying Eq. (70) by itself,

 

$$\lim_{n\to\infty} F^{2n} (a_n x + b_n) = \psi^2_a(x) = \begi... ...& \text{for } x < 0, \\ 1, & \text{for } x \geq 0. \end{cases}$$ (72)

Set $\gamma = 2^{1/a}$. Then, Eq. (72) becomes

 

$$\lim_{n\to\infty} F^{2n} (a_nx + b_n) = \psi_a (\gamma x).$$ (73)

The transformation $z = \gamma x$ changes Eq. (73) to

 

$$\lim_{n\to\infty} F^{2n} \left( \frac{a_n z}{\gamma} + b_n \right) = \psi_a (z).$$ (74)

Now, replace z by x in Eq. (74), since the relationship will still be true after such a replacement. Then,  Eq. (74) becomes

 

$$\lim_{n\to\infty} F^{2n} \left( \frac{a_n x}\gamma + b_n \right) = \psi_a (x).$$ (75)

Equations (71) and (75), together with Theorem 2, require that

$$\lim_{n\to\infty} \frac{a_{2n}}{\frac{a_n}\gamma} = 1 \quad ... ...ad \lim_{n\to\infty} \frac{b_{2n} - b_n}{\frac{a_n}\gamma} = 0.$$

This condition will be satisfied if

 

$$a_{2n} = \frac{a_n}\gamma \quad \text{and} \quad b_{2n} = b_n.$$ (76)

Hence, by Theorem 2, the use of Eq. (76) to generate a new set of constants is permissible. The new set will still satisfy Eq. (71). Call the new set a'_k and b'_k, where

 

$$b'_k = b_{n2^k} \quad \text{and} \quad a'_k = \frac{a_n}{\gamma^k},$$ (77)

n being some fixed integer. But, by Eq. (76),

 

$$\begin{array} {r@{~}c@{~}l} b_{n2^k} & = & b_{n2^{k-1}} = \l... ...e \frac{a_{n2^{k-1}}}\gamma = \frac{a_n}{\gamma^k}. \end{array}$$ (78)

Then,

 

$$\lim_{k\to\infty} a'_k = \lim_{k\to\infty} \frac{a_n}{\gamma^k} = 0, \quad \text{since } \gamma = 2^{1/a} \gt 1.$$ (79)

Furthermore, a'_k and b'_k still satisfy Eq. (71). Then, for x = 0, b'_n must be such that

 

$$\lim_{k \to\infty} F^k (b'_k) = \psi_a (0) = 1.$$ (80)

If F(x) < 1 for all finite x, then $b'_k \to \infty$ as $k \to\infty$ or by Eq. (78) all b'_k are infinite. This contradicts that $F^n(a_n x + b_n) \to \psi_a (x)$, since a_n > 0, and b_n must be finite (at least for finite n) if Fⁿ(a_n x + b_n) approaches a proper limiting frequency distribution. Hence, there exists x₀ such that F(x₀) = 1 and F(x) < 1 for x < x₀. This demonstrates condition (a) of Theorem 14.

In the above development, b'_k could arbitrarily be assigned the value x₀ for all $k = 1,2,3, \ldots$ without affecting any of the argument. Suppose that this assignment is made. For convenience the k in a'_k and b'_k will be replaced by n and the primes dropped since this sequence of constants could have been chosen in the first place without affecting any of the discussion. Then, Eq. (70) requires that

 

$$\lim_{n\to\infty} F^n (a_n x + x_0) = \begin{cases} e^{-(-x)... ... & \text{for } x < 0, \\ 1, & \text{for } x \geq 0 \end{cases}$$ (81)

or

 

$$\lim_{n\to\infty} n [ 1 - F (a_n x + x_0)] = e^{-(-x)^a}, \qquad \text{for } x < 0.$$ (82)

For all y < 0 sufficiently close to zero, it is possible to find an n value such that either

 

$$-a_n \leq y \leq -a_{n+1} \quad (\text{if } a_{n+1} \leq a_n)$$ (83)

or

 

$$-a_{n+1} \leq y \leq -a_n \quad (\text{if } a_n \leq a_{n+1}).$$ (84)

If Eq. (83) holds,

 

$$1-F (-a_{n+1} + x_0) \leq 1 - F (y+x_0) \leq 1 - F (-a_n + x_0).$$ (85)

Also, for all k > 0,

 

$$1 - F (-ka_{n+1} + x_0) \leq 1 - F (ky + x_0) \leq 1 - F (-ka_n + x_0).$$ (86)

From Eqs. (85) and (86),

$$\frac{1 - F (-ka_{n+1} + x_0)}{1 - F (-a_n +x_0)} \leq \frac... ... x_0)} \leq \frac{1 - F (-ka_n + x_0)}{1 - F (-a_{n+1} + x_0)}$$

or

$$\frac{n}{n+1} \ \frac{(n+1) [ 1 - F (-ka_{n+1} + x_0)]}{n[1-... ... \ \frac{n [1-F (-ka_n + x_0)]}{(n+1) [ 1-F (-a_{n+1} + x_0)]}.$$

Hence, as $n \to \infty$, by Eq. (82), since $\lim_{n\to\infty} (n/(n+1)) = 1$,

$$(+k)^{1/a} \leq \frac{1-F (ky+x_0)}{1-F (y+x_0)} \leq (+k)^{1/a}$$

or

 

$$\lim_{n\to\infty} \frac{1-F (ky+x_0)}{1-F (y+x_0)} = (+k)^{1/a}.$$ (87)

But, by Eq. (83), $y \to -0$ as $n \to \infty$,because $-a_n \to -0$. So, Eq. (87) can be restated as

$$\lim_{y\to-\infty} \frac{1 - F (ky+x_0)}{1 - F (y+x_0)} = (+k)^{1/a}.$$

If Eq. (84) holds, an exactly parallel line of reasoning leads to the same result. This proves condition (b) of Theorem 14.